If the equation of projection is #y=ax-bx^2#, find initial velocity(#u#) and angle of projection(#theta#)?

2 Answers
Jun 25, 2017

Given that the equation of projection is #y=ax-bx^2#, we are to find initial velocity(#u#) and angle of projection(#theta#)

Horizontal component of velocity of projection is #ucostheta#

Vertical component of velocity of projection is #usintheta#

Let at the t th sec after its projection from origin its position be represented by the coordinates #(x,y)#

So #x=ucosthetaxxt#

#=>t=x/(ucostheta).....[1]#

Again

#y=usinthetaxxt-1/2gxxt^2.....[2]#

Combining [1] and[2] we get

#y=usinthetaxxx/(ucostheta)-1/2gxx x^2/(u^2cos^2theta)#

#=>y=xtantheta- (gx^2)/(2u^2cos^2theta)....[3]#

Comparing equation [3] with the given equation we get

#a=tantheta#

#=>theta=tan^-1a#

and

#b=g/(2u^2cos^2theta)#

#=>u^2=g/(2b)sec^2theta#

#=>u^2=g/(2b)(1+tan^2theta)#

#=>u^2=g/(2b)(1+a^2)#

#=>u=sqrt(g/(2b)(1+a^2))#

Jun 25, 2017

See below.

Explanation:

This is a typical kinematic problem. Assuming that the movement begins at the referential origin we have

#{(x = v_x t),(y=v_y t-1/2 g t^2):}#

where

#t# is the elapsed time
#v_x# the #x#-axis velocity component
#v_y# the #y#-axis velocity component
#g# the gravity acceleration.

Now the orbit equation is

#y-ax+bx^2=0# substituting the parametric equations we have

#1/2(g t + 2 a v_x - 2 b t v_x^2 - 2 v_y)t=0#

This condition must be true for all #t# so

#(g t + 2 a v_x - 2 b t v_x^2 - 2 v_y)=0, forall t#

Now grouping coefficients

#(g - 2 b v_x^2)t+2(a v_x - v_y)=0# so

#{(g - 2 b v_x^2=0),(a v_x - v_y=0):}#

Now solving for #v_x, v_y#

#{(v_x=sqrt(g/(2b))),(v_y=a sqrt(g/(2b))):}#

Now #v=sqrt(v_x^2+v_y^2) = sqrt((g/2)((a^2+1)/b))#

#theta = arctan(v_y/v_x) =arctan(a)#